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The  Compression  and  Transmission 
of  Illuminating  Gas 


A  Thesis  Read  at  the  July,  1905,  Meeting  of  the  Pacific  Coast 
Gas  Association 


By  E.  A.  Rix 


Mem.  Am.  Soc.  Mech.  Eng.      Mem.  Am.  Soc.  Civil  Eng.      Assoc.  Mem.  Am.  Soc 
Mining  Eng.      Mem.  Pacific  Coast  Gas  Assn. 


San  Francisco,  Septembrr  i,  1^5. 

The  Journal  of  Electricity  Publishing  Company 
,  Room  S32,  Rialto  Building 
San  Francisco,  Cat. 


^z- 


THE    COMPRESSION   AND    TRANSMISSION  OF  ILLUMI- 
NATING GAS. 

The  subject  of  illuminating  gas  compression  is  almost 
a  new  one,  and  the  nature  of  the  gas  is  so  entirely 
different  from  that  of  air  that  we  are  obliged  to  consider 
the  question  mainly  from  the  theoretical  standpoint, 
backed  up  by  a  few  indicator  cards,  which  have  been 
furnished  us  by  gas  compressors.  But  you  may  be 
assured  that  all  of  the  data  given  herewith  is  eminently 
practical,  because  there  has  been  eliminated  all  of  the 
small  variables  that  are  important  fiom  a  chemical  stand- 
point, but  which  the  advancing  piston  of  a  compressor 
cylinder  takes  little  heed  of. 

We  are  not  concerned  about  the  candle  power  or  the 
commercial  utility  of  a  gas,  but  simply  with  its  weight 
and  composition,  and  what  may  happen  to  it  after  it 
leaves  the  compressor  cylinder  is  not  the  province  of  this 
paper. 

All  gases  are  sponge  like  in  that  they  hold  various 
vapors  from  water  vapor  to  carbon  vapors,  which  they 
lose  to  a  more  or  less  extent  when  the  sponge  is  squeezed 
as  in  the  act  of  compressing  in  a  cylinder,  and  what  is 
squeezed  out  and  how  much  of  it  is  not  essential  to  our 
discussion,  and  lies  better  in  the  realm  of  the  technical 
gas  engineer. 

We  have  assumed,  however,  that  inasmuch  as  when  we 
compress  a  gas  the  temperature  rises  in  a  fixed  ratio  to 
the  pressures,  that  there  is  no  direct  tendency  for  a  gas 
to  change  its  physical  condition  in  the  compressing 
cylinder,  for  an  added  temperature  gives  an  added 
capacity  for  saturation,  and  this  probably  increases  in 
about  the  same  ratio  as  the  volume  diminishes  during 
compression.  So  that  for  commercial  purposes  we  cannot 
be  far  wrong  in  assuming  the  physical  condition  of  the 
gases  as  constant  during  the  range  of  pressures  that  will 
be  ordinarily  met. 

All  phenomena  of  compression  and  expansion  of  gases 
is   intimately   associated  with  temperature,  in  fact  the 


.517345 


4  THE  COMPRESSION   AND  TRANSMISSION   OF  GAS. 

power  to  compress  any  gas  in  foot-pounds  is  simply  the 
difference  in  temperature  between  the  gas  before  and 
after  compression,  multiplied  by  its  weight  in  pounds, 
by  its  specific  heat,  and  then  by  Joules  equivalent  to  con- 
vert heat  units  to  foot-pounds.  Expressed  algebraically, 
this  equation  is: 

^=/f^<:'p(^-  ^o)  where 

y  is  Joules  equivalent  =  772 

W  =  the  weight  in  pounds  avoirdupois  to  be  com- 
pressed. 

Cp  is  the  specific  heat  of  the  gas  at  constant  pressure. 

T^  is  the  initial  absolute  temperature. 

7"  is  the  final  absolute  temperature. 

L  is  the  work  expressed  in  footpounds. 

This  is  the  general  equation  for  the  compression  of 
any  gas. 

In  glancing  at  this  equation,  the  first  stumbling  block 
we  strike  is  C,,  the  specific  heat  of  the  gas  at  constant 
pressure,  and  this  must  be  first  determined.  After  that, 
we  must  discover  some  means  of  finding  T  the  final 
temperature. 

To  anticipate  a  little,  it  may  be  stated  here  that  these 
temperatures  are  all  functions  of  the  ratio  of  the  specific 
heats  of  gas  at  constant  pressure,  and  at  constant  volumes. 

It  is  then  our  first  duty  to  understand  about  these  two 
specific  heats  and  to  know  how  to  determine  them  for 
any  gas,  and  the  rest  is  simple. 

The  specific  heat  of  any  substance  is  the  amount  of 
heat  one  pound  of  that  substance  will  absorb  to  raise  its 
temperature  1°  Fah.,  the  specific  heat  of  water  being  i. 

When  a  gas  is  heated  two  different  results  may  be 
obtained,  depending  upon  whether  the  gas  is  allowed  to 
expand  and  increase  its  volume  when  heated,  the  pres- 
sure remaining  constant,  or  whether  the  air  is  corfined 
and  the  volume  remain  constant,  and  the  pressure  in- 
creasing. The  amount  of  heat  to  raise  the  temperature 
of  a  gas  1°  under  these  two  conditions  is  different,  there- 
fore, the  specific  heat  is  different.  The  former  is  called — 
Specific  heat  at  constant  pressure,  and  the  latter— Speci- 
fic heat  at  constant  volume. 


THE  COMPRESSION   AND  TRANSMISSION   OF  GAS. 


TABLt:  J. 


July  J305 


/SOTH£/iliAL  CUKYE  /'oVe-  fl^"  CONS 'T 


/iOIABATIC  Cu/f  V£     /i  Vo 


'  Z'/" 


'COMS'T. 


/-joJ 


/^G-jf      I 


6  THE   COMPRESSION    AND   TRANSMISSION   OF   GAS. 

Referring  to  Table  i,  Figure  3,  if  we  have  a  cylinder  A, 
containing  one  pound  of  gas  at  atmospheric  pressure, 
and  a  piston  P,  without  weight,  but  having  an  area  of 
one  square  foot,  and  heat  the  gas  until  the  temperature 
has  risen  1°  Fah,,  the  gas  will  have  expanded  by  the 
small  amount  d,  as  in  Figure  4,  and  raised  the  piston. 
This  expansion  is  1/460  of  the  original  volume,  at  0°  Fah. 

It  is  evident  that  inasmuch  as  the  piston  has  raised 
and  displaced  the  atmosphere,  that  work  has  been  done, 
which  must  have  absorbed  heat  in  addition  to  that  neces- 
sary to  raise  the  temperature  of  the  air  1°,  If  the  pis- 
ton was  fastened,  as  in  Figure  3,  the  gas  would  have 
required  just  that  less  heat  to  raise  it  1°  as  was  reqnired 
to  lift  the  piston  through  the  distance  d  =  1/460  of  its 
volume.  The  amount  of  heat  required  in  the  first  in- 
stance is  called  specific  heat  at  constant  pressure,  and 
the  latter  at  constant  volume. 

Specific  heat  of  most  of  the  gases  at  constant  pressure 
has  been  determined  by  Regnault  and  others  experiment- 
ally, and  the  symbol  is  C^. 

The  amount  of  work  done  in  lifting  the  piston  through 
the  distance  d  is  measured  the  same  as  the  work  done  by 
any  piston  by  multiplying  the  pressure  on  the  piston 
by  the  distance  passed  through.  The  area  multiplied  by 
the  distance  is  the  volume,  which  may  be  expressed  by 
V.  The  distance  d  \s  1/460  at  0°  Fah.,  or  may  be  ex- 
pressed by  y^ 

lyCt  P  be  the  pressure,  and  P,  the  foot-pounds  of  work 
dene,  then 

VP 

—=r  =  R  and  this  is  called  the  Simple  Gas  Equa- 
tion, and  about  it  hangs  many  important  deductions. 

i?  is  a  constant  for  any  gas,  because  inasmuch  as  the 
gas  expands  uniformly  for  each  1°  of  heat,  any  volume 
as  Fj  multiplied  by  its  corresponding  P^  and  divided  by 
its  corresponding  temperature  Z)  will  equal  R,  or  to  put 
it  algebraically, 

VP       V,  P,         V"  P" 


T  7;  T" 


=  ^  =  Constant 


THE  COMPRESSION   AND  TRANSMISSION   OK  GAS. 


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8  THE   COMPRESSION   AND  TRANSMISSION  OF  GAS. 

R  being  always  in  foot-pounds,  if  we  divide  it  by 
Joules  equivalent  772,  which  is,  as  you  know,  the  amount 
of  foot-pounds  equal  to  i  heat  unit,  and  which  is  always 
denoted  by  /,  we  shall  have  the  amount  of  heat  units 
that  were  converted  into  work  to  raise  the  piston,  and 
this  amount  of  heat,  we  know,  must  be  the  difference 
between  the  specific  heat  at  the  constant  pressure  and 
the  specific  heat  at  constant  volume,  or, 

y      '     ' 

from  which  we  have 

C  =C  — - 
\      "       / 

an  equation  from  which  the  specific  heat   at   constant 

volume  may  be  determined  for  any  gas  within  the  limits 

of  its   stability,    and   certainly    within    the   commercial 

pressures  you  are  likely  to  encounter. 

For  a  perfect  gas,  these  specific  heats  are  practically 

constant;    that  is,   they  are  not  affected  by  pressure  or 

temperature,  but  so  far  hydrogen   and   air  appear  to  be 

nearer  than  any  other  gases.     CO  and  CO.^,   which  are 

inferior  components  of  illuminating  gas,    as  it  is  now 

made,  shows  the  greatest  deviation,   but  not  enough  to 

render  their  vagaries  of  moment  in  the  consideration   of 

the  power  question,  consequently  all  the  following  data 

has  been  calculated  on  the  basis  of  the  simple  gas  law. 

P  V 
„  =  R  =  Constant. 

As  an  example  showing  how  to  calculate  the  specific 
heat  at  constant  volume,  let  us  take  C.^  H^.  This  gas  has 
been  selected  because  of  an  evident  error  in  the  values 
ascribed  to  Regnault  in  the  references  we  have  at  hand. 

Upon  applying  the  simple  gas  equation  to  the  Reg- 
nault value  there  was  a  large  discrepancy,  and  it  will  be 
interesting  no  doubt  to  make  the  calculations  here,  and 
thus  make  them  serve  the  double  purpose  of  showing 
how  to  determine  the  specific  heat  at  constant  volume 
and  to  point  out  the  error. 

Regnault  gives  the  C^,  of  Q  H^  to  be  .404,  and  C^  to 
be.  1 73.  The  weight  per  cubic  foot  to  be  .0780922,  or 
12.8  cubic  feet  in  one  pound  at  32^  Fah. 


THE   COMPRESSION    AND  TRANSMISSION  OF   GAS. 


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lo  THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 

If,  now,  one  pound,  or  22.30  cubic  feet,   be  heated  to 

1°  Fah.  and  allowed  to  expand,  the  simple  gas  equation 

P  V 

— ™^  =  R  will  give  at  32° 

22.39  X   144  X   12.8  _ 

492  ^^ 

Fifty-five  foot-pounds  of  work  has  been  performed  by 
the  gas  in  expanding  against  the  atmosphere;  to  convert 
this  into  heat  units  we  divide  by  Joules  equivalent  772. 

—  =  .07124  units  of  heat. 
772 

Inasmuch  as 

C^  =  C|,  —  —  and  —  =  .07124 

we  have  C^  =  .404  —  .07124  =  .3327,  instead  of  .173  as 
determined  by  Regnault.  The  ratio  between  the  two 
specific  heats  forms  the  basis  for  all  the  calculations  for 
the  relations  between  pressure,  volume  and  temperature 
in  compressing  gas,  and  that  is  why  we  must  be  par- 
ticular about  these  specific  heat  factors. 

C 

-^-  =  y,  which   we  shall  discuss  further  on,   and 

which  is  brought  in    now    simply   as   additional    proof 

about  the  figures  which  we  have  just  obtained  for  C.^  H^. 

For  C.^  //<,  using  Regnault's  values,  we  have 

.404 
y=~TZZ=  2-33 

•  173 

for  our  values 

.404 

r  = =  1. 2 14. 

.3327 
In  reading  a  new  book  by  Travers  on  the  study  of 
gases  (page  275),  he  gives  some  very  interesting  calcu- 

C 
lations  to  show  the  limiting  values  of     "  ox  y. 

His  conclusions  are  that  for  a  monoatomic  gas  within 

PV 
the  limits  of  the  simple  gas  equation  -^„    =  R,    he  val- 

C 
ues  of -^t  can  never  exceed   1.667,    ^^^  the  value  for  a 

diatomic  gas  should  range  about  1.4  and  the  polyatomic 
gases  still  less,  until  we  reach  the  value  of  i,    where,   of 


THK   COMPRESSION    AND   TRANSMISSION    OK   GAS. 


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12  THE   COMPRESSION   AND   TRANSMISSION   OF   GAS. 

course,  there  should  be  no  expansion  work  at  all  when 

heat  was  applied. 

C 
We  can  see,  therefore,  that  the  value  -^  of  2.33  from 

Regnault's  values  is  an  impossibility,  the  maximum  pos- 
sible value  being  only  1.667,  and  Q  H^  being  the  poly- 
atomic gas,  its  value  would  be  less  than  1.4,  all  of  which 

.      .  C 

indicates  that  our  figures  -^  =^  1.2 14  are  approximately 

correct. 

It  will  now  be  necessary  to  apply  our  understanding 
of  these  principles  and  try  and  determine  the  values  of 
the  specific  heats  for  illuminating  gas.  There  seems  to 
be  plenty  of  data  about  the  specific  heat  at  constant  pres- 
sure for  gas  mixtures,  but  nothing  about  the  specific  heat 
at  constant  volume. 

Reference  is  now  made  to  the  Tables  2,  3,  4,  5,  6,  7 
and  8,  which  show  the  composition  and  heat  properties 
of  seven  different  gases  and  the  methods  employed  in 
determining  the  weights,  specific  gravities  and  specific 
heats. 

Column  I  is  the  chemical  symbol  for  the  different  com- 
ponents. 

Column  2  is  the  percentage  by  volume  of  the  different 
components. 

Column  3  gives  reliable  weights  per  cubic  foot. 

Column  4  gives  the  specific  heat  of  each  component 
gas  as  determined  by  Regnault  and  others. 

Column  5  gives  the  productof  the  different  percentages 
of  the  component  gases  and  their  weights  per  cubic  foot, 
or  Column  2  multiplied  by  Column  3.  The  total  sum 
divided  by  100  gives  the  weight  of  the  gas  per  cubic  foot. 

Column  6  gives  the  product  of  Column  4  and  Column 
5  for  specific  heat,  being  a  weight  function.  We  must, 
in  order  to  get  the  specific  heat  of  the  compound  gas, 
take  into  consideration  not  only  the  percentages  of  the 
component  parts,  but  the  weights  as  well,  and  also  the 
specific  heat  of  each  component.  The  sum  of  the  products 
in  column  divided  by  100,  and  then  by  the  weight  of  one 
cubic  foot  of  the  compound  gas,  will  give  the  specific 
heat  at  constant  pressure  C^. 


THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 


13 


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14 


THE   COMPRESSION   AND   TRANSMISSION   OF   GAS. 


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THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 


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THE  COMPRESSION    AND   TRANSMISSION   OF   GAS. 


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THE  COMPRESSION   AND  TRANSMISSION   OF  GAS. 


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i8  The  compression  and  transmission  of  gas. 

Column  7   gives  the  calculations  to  find  the  specific 

C 
heat  at  constant  volume  and  also  R  and  ~  or  y  for  each 

gas,   and   also   various   factors  of  y  which  we  will  find 
useful  later. 

Table  9  concentrates  Tables  2  to  8,  so  that  we  may 
study  them  easier. 

You  will  note  that  our  results  cover  quite  a  field, 
taking  in  California  fuel  oil  gas,  Massachusetts  coal  gas, 
Indiana  natural  gas,  California  natural  gas,  and  Cali- 
fornia carburetted  water  gas,  and  after  carefully  study- 
ing their  heat  and  power  properties,  as  shown  in 
Table  9,  we  have  selected  the  fuel  oil  gas  made  in  Oakland 
as  having  the  best  average  properties  for  the  purposes 
we  have  in  view,  and  particularly  as  fuel  oil  gas  is  the 
one  you  will  probably  have  most  to  deal  with. 

We  may  therefore  consider  our  subject  as  having  for  a 
basis  a  gas  with  the  following  properties  at  32°  Fah 
Weight  per  cubic  foot,  .0323577, 
Cubic  foot  in  one  pound  avoirdupois,  30.98. 
Specific  gravity,  .4008. 

c;  =  .6884. 

y  ^  I-334- 
y-i 


y 
y 


=  .25. 


y-^    ='• 

R=  133.2. 

L  =  8467  (^„  =  I  ) 

A  cubic  foot  of  gas  varies  in  weight  according  to  the 
altitude  or  pressure,  and  also  according  to  the  temper- 
atures.    The  law  of  this  variation  is  expressed  as  follows: 

Having  given  the  weight  of  a  gas  for  any  temperature, 
or  any  pressure,  then  the  weight  at  any  other  temper- 
ature or  pressure  will  be  as  the  ratio  of  absolute  temper- 
ature or  pressure,  or 

W  =  W  ~  or  W~  where 

IV=  known  weight. 


THE   COMPRESSION   AND  TRANSMISSION  OF  GAS.  I9 

T"  and  P"  the  known  temperature  or  pressure  and  W 
the  desired  weight. 

For  example — Our  standard  gas  weights  at  sea  level, 
or  14.7  pounds  absolute  pressure,  and  32°  Fah.,  .03235 
pounds  per  cubic  foot;  at  20  pounds  gauge,  or  34.17 
pounds  absolute,   a  cubic  foot  would    weigh    .03235  X 

— --  =  .03235  X   2.36  =   .076346  pounds,   and  at  60° 
14.7 

Fah.,  instead  of  32°  Fah.,  this  cubic  foot  would   weigh 

520 
.076346  X   —-  =  .0819  pounds.     460  being  the  absolute 
460 

temperature  of  O"  and  520  the  absolute  temperature  of 

60°  Fah.  =  460  +  60  =  520. 

Altitudes  are  nothing  more  or  less  than  pressures  less 
than  sea  level,  and  are  treated  just  the  same  as  pressures 
above  the  normal  atmospheric. 

Thus  at  5225  feet  the  absolute  pressure  is  12.044,  con- 
sequently, as  gas  at  this  altitude  would  weigh   — '_z^ 

times  the  weight  at  sea  level. 

For  your  convenience  it  may  be  well  to  add  here  that 
when  the  barometric  pressure  is  knovyn,  the  atmospheric 
pressure  is  found  by  multiplying  the  barometric  pressure 
by  .4908,  or  P"  =^  B  X  .4908. 

For  example — When  the  barometric  is  29.92  the 
atmospheric  pressurre  is  29.92  X  4908.  or  14,7,  the  nor- 
mal sea  level  pressure. 

To  find  the  atmospheric  pressure  when  the  altitude  in 

feet  is  given,  we  have 

57000  A^ —  N'^  .       ,  .  , 

P"  =  14.72  —  in  which 

100,000,000 

A^  =  altitude  in  feet. 

For   example — To   find   the    atmospheric   pressure    a 

10,000  feet  we  have  

57,000  X   10,000  v<   (10,000)'^ 

P"  =  14.72 —  or 

100,000,000 

P"  =^  1472  —  4.7  =  10.02,  the  atmospheric  pres- 
sure required. 

The  foregoing  rules  will  be  all  that  is  necessary  to  calcu- 
late all  variations  of  weights  due  to  pressure,   altitud*' 


20  THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 

or  temperature,  and  relative  volumes  follow  exactly  the 

same  laws  as  relative  weights. 

For   convenience   in    many   calculations   Table    lo    is 

P 
given  herewith,  showing  the  pressure  ratios,  or  —  for 

every  pound  from  i  to  no,  and  the  volumes  ratios  will 
be  inversely  as  the  pressure  ratios  and  consequently  the 
reciprocal  of  the  figures  on  the  table. 

This  might  be  called  a  table  showing  also  the  rates  of 
Isothermal  compression  or  expansion  or  Marriotte's  law, 
the  general  formula  for  which  is: 

po  yo  ^  p  y_  Constant,  or  in  other  words,  the  pro- 
duct of  any  pressure  by  its  volume  is  always  equal  to  the 
product  of  any  other  pressure  by  its  volume,  and  this 
rule  will  be  found  useful  in  determining  the  contents  of 
receivers,  etc.  It  must  always  be  remembered  that  in 
using  these  rules  all  temperatures  must  be  alike,  or  cor- 
rections made  according  to  the  rules  just  given. 

ISOTHERMAL    COMPRESSION. 

There  are  two  methods  of  compressing  any  gas. 

First — Where  the  temperature  remains  unchanged 
during  compression.  This  is  called  Isothermal  compres- 
sion and  is  the  ideal  method  never  realized  in  practice. 

Second — Adiabatic  compression,  which  is  the  kind  we 
meet  in  practice  where  the  heat  developed  by  compression 
expands  the  air  being  compressed  until  it  follows  a  dif- 
ferent law  from  Marriotte. 

While  Isothermal  compression  is  not  practical,  it  is 
necessary  to  know  about  it  and  how  to  make  the  calcul- 
ations concerning  it. 

We  have  found  that  the  volume  ratios  are  inversely  as 
the  absolute  pressure  ratios  in  Isothermal  compression. 
Consequently  if  the  pressure  ratios  are  i,  2,  3  and  4,  ihe 
corresponding  volumes  will  be  i,  }4,  yi,  }(.  To  show 
this  graphically — refer  to  Table  i.  Figure  i. 

Let  A  B  he  the  line  of  o  pressure  or  the  perfect  vacuum 
line.  C  D  the  intake  line  and  we  erect  pressures  ordi- 
nates  G  H  =  2  X  Z>i9ata  point  //"equal  to  ^ ,  ^  ^  and 
I  J—  ^  X  B  D  2X2.  point  J  =  Yz  A  B  and  if  A'  =  4  x 
B  D  2iX.  sl  point  K  =  %  oi  A  B  counting  all  volumes 
from  F  B  or  the  end  of  the  piston  stroke. 


THE  COMPRESSION    AND   TRANSMISSION   OF  GAS.  21 

If  we  join  the  points  C  G  I E  \n  a.  curved  line,  it  will 
be  the  Isothermal  or  logarithmic  curve  and  it  will  be 
noted  that  the  area 

E  FB  K=  4  X    14:  =  I 
ILBJ^2>  X  /3  =  1 
G  M B  H=  2  X   >^  =  I 
CDBA^\y.\Qx\ 
As  found  before.  /""  V  =  P'  V  =  Constant,  and  the 
figure  represents  the  ideal  indicator  card  for  Isothermal 
compression     for   four   compressions,   counting  from    o, 
and  the  above  method  will  always  be  proper  to  lay   out 
an  Isothermal  curve,  no  matter  what  the  intake  pressure 
may  be. 

To  find  the  work  of  compression  and  delivery  Isother- 
mally 

d^^P"  V  hyp.  log  ^  in  foot  pounds  in  which 

P"  =  Initial  pressure  absolute. 
V  =Initial  Volume 
P  =  Final  pressure 
L  =  Work  required. 
In  all  of  our  calculations  V°  will  be  taken  as  one  cubic 
foot. 

For  Example — Haw  many  foot-pounds  of  work  is  re- 
quired to  compress  i  cubic  foot  of  gas  at  sea  level  to 
eighty  pounds  gauge  pressure. 

For  sea  level  P"  per  square  foot  =  14.7  x  144  = 
2116.8  pounds.     Then 

p 
L=  21 16.8  hyp.  log.  ^„ 

Consulting  Table  10  we  find 

p 

--  for  80  pounds  gauge  =  6.442  the  hyperbolic 

logarithm  of  which  is  1.863. 

Substituting,  we  have 

L  =  2116.8  X   1.863  =  3943  foot-pounds. 

If  a  table  of  hyperbolic  logarithms  is  not  at  hand,  it 
would  be  well  to  remember  that  hyp,  log.  =  common 
log.  X   2,3026. 


THE  COMPRESSION   AND   TRANSMISSION   OF  GAS. 


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THE   COMPRESSION   AND   TRANSMISSION   OF  GAS.  23 

The  HP.  required  for  above  work  will  be  -3945  _ 

33000 

•  1195  ffP 

To  find  the  M E  P  oi  Isothermal  compression, 

P 
M  E  P  =^  P"  hyp.  log.  —  ,  using  the   quantities 

in  the  previous  example  we    have  M  E  P  =^   i^."]    X 

1.863    =    27.38    pounds.       We     know    that     HP.      = 

ME  P  X   V 

and  for  one  cubic  foot  V  =  i  x  144.     Con- 

33000 

sequently,  using  the  last  example, 

rrr^  27.38    X     I     X      I44  ^, 

HP  = -^—^ ~~  =   .1195,  the  same  re- 

33000 

suit  as  before.     —  =     .00436.      Consequently     a 

33000 

short  and  convenient  formula  would  be  for  Isothermal 
compression  HP=  .00436  X  M  E  P. 

It  will  be  noted  that  none  of  the  physical  properties  of 
gases  enter  into  the  above  equations,  consequently  we 
must  conclude  that  it  takes  the  same  power  to  compress 
one  cubic  foot  of  any  gas  Isothermally  to  the  same  pres- 
sure, provided  the  ratios  of  pressures  are  the  same. 

ADIABATIC  COMPRESSION. 
We  have  before  stated  that  Isothermal  compression  is 
ideal,  and  not  realized  in  practice.  All  of  the  work  ex- 
pended in  compressing  a  gas  is  converted  into  heat  in- 
stantly, and  this  increases  both  the  temperature  and  the 
volume  of  the  gas  during  compression,  so  that,  in- 
stead of  having  a  relation  between  pressure  and  volume 
(/>^  J7,,  =  />  J^  =  Constant),  such  as  we  found  in  Isoth- 
ermal compression,  we  now  have  a  relation  /*„  V J  = 
P  V^  =  Constant,  or  in  other  words,  the  gamma  powers 
of  each  volume,  multiplied  by  its  corresponding  pressure, 
is  Constant.  This  is  the  equation  of  the  Adiabatic 
curve,  y  is  the  same  that  we  found  to  be  the  ratio  be- 
tween the  specific  heat  at  constant  pressure  and  that  at 
constant  volume.  This  relation  can  perhaps  be  fastened 
a  little  easier  in  the  mind  by  remembering  that  the  equa- 
tion of  the  Isothermal  curve  represents  the  law  of  Mar- 
riotte  and  the  equation  of  the  adiabatic  curve  represents 
the  Exponential  law  of  Marriotte. 


24  THE  COMPRESSION   AND  TRANSMISSION   OF  GAS. 

Inasmuch  as  the  power  to  compress  a  gas  is  measured 
practically  by  the  indicator  diagram,  and  this  in  turn  is 
compared  to  the  adiabatic  curve  which  is  theoretical 
curve  of  compression,  and  inasmuch  as  we  depend  upon 
the  value  of  y  to  construct  this  curve,  it  will  be  at  once 
seen  why  we  were  to  particular  to  discover  the  relation 

— -  =  jj/.     Now  if  Po  Vj  =  P  V-'  and  from  the  single  gas 

yo  po        y  p 
equation -^^3— =  "-™-   =    R,    by  combining    these    we 

have  all  the  adiabatic  relations  between  volume  pressure 
and  temperature  as  follows: 

T"    \v )      yp" ) 


It  will  always  be  necessary  to  use  the  above  formulae 
in  making  calculations  for  pressures,  temperatures  and 
volumes,  or  for  power  to  compress  any  gas  which  varies 
far  enough  from  the  standard  we  have  selected  to  make 
it  necessary,  but  there  is  no  doubt  that  for  all  practical 
purposes,  at  least  for  the  present,  Table  ii,  which  is 
calculated  for  our  standard  gas,  will  give  the  proper 
values  for  rapidly  and  easily  calculating  any  problems 
connected  with  compressing  illuminating  gas. 

All  reference  to  expansion  is  purposely  omitted,  be- 
cause gas  will  probably  never  be  used  for  expansion 
work  in  an  engine  as  air  is  used 

Assuming  that  all  may  not  be  familiar  with  just  how 
to  arrive  at  the  results  as  indicated  in  Table    1 1 ,  let  us 

P 

take  a  ra.ioof -7^^=  2  corresponding  to  14.7  pounds  gauge 

V  T 

pressure  and  discover  what  are  the  values  of  7^  and  -=i, 

V        /P\~ 
we  have  7^  =  I  — ^  j    y    y  we  have  already  decided  from 

uor  standard  gas  to  be  1.334. 


THE  COMPRESSION   AND  TRANSMISSION    OF   GAS.  25 

II  V  /P°\    "'** 

Therefore,  —  = = .  749    775-=  (  -77  )  or 


since 


-77  =  ~  or  .  5  we  nave 
±*        2 

V 

-p=.5""   or 

Log   y5  =  log  .5  X  .749 

Log      .5    -^    1.6989     X     .749  =   1-77447    =    log 

V  V  v 

-p^  giving  value  of -r?5  =.  .5949   and  77-  will  be  recipro- 

V 
cal   of  „„   or  1. 68 1. 
r 

To  find  the  ratio  of  temperature  for   this  same  rate 

c  •  ^         T     /P\'^      y-i 

of  compression,  we  have  -~^  =(  ^^j  y        - —  =  .25. 

Hence: 

T  P 

Log  y„z=.25log.  ^ 

P  T 

Log.  p^=  .301    X    .25  =    .07525   =    Log.  ^ 

-f-^  =  -1892 

T  =  520  X  1-1892  =  618°  absolute  or  158°  Fah. 
If    T°  =  60°  Fah.     We  have  then 

P  V  V 

—  =  2      -pr=  1.681    p:^=.5949 

^=1.1892  T=  158° 

Air  under  the  same  conditions  gives 

P  V'^         ^  V        ^ 

po  =  2        -p-=  1.6349  Yo  =  -^"7 

T 
y-  =  1.2226  T  =  175°  Fah 

These  examples  will  serve  to  show  how  this  Table  1 1 
was  calculated.     A  few  examples  will  show  its  use. 

Problem — To  find  the  final  temperature  due  to  adiaba- 
tic  compreSvSion. 


26  THE  COMPRESSION    AND  TRANSMISSION  OF  GAS. 

P  T 

Opposite    —  and  under  the  headline  —  will  be  found 

the  ratio  of  absolute  temperatures. 

Example — What  is  the  final  temperature  due  to  14.7 
pounds  gauge  pressure  at  sea  level  and  60°  Fah. 

-=  =  — —  =2.     Then  -^^  =  1.1892,  or. 520  x  1.1892 
/^,       14.7  1 

=  618°  abs.  or  158°  Fah. 

If  the  initial  temperature  has  been  100°  then  560  x 
1. 1892  z.  666°  Fah. 

It  is  readily  noted  from  this  that  the  higher  the  initial 
temperature,  the  higher  the  final  temperature,  and  it  will 
also  be  noted  that  while  there  is  a  difference  of  40°  be- 
tween the  initial  temperature,  there  is  a  difference  to  48° 
between  the  final  temperatures;  a  difference  of  8°. 

Iiiasm  I  :h  as  the  temperature  developed  during  com- 
pression is  at  the  expense  of  power,  it  is  evident  that  it 
takes  more  power  to  compress  the  same  weight  of  gas  at 
100°  Fah.  then  at  60°  Fah.  to  the  same  pressure,  all 
other  conditions  being  similar. 

It  is  an  axiom,  therefore,  that  the  cost  of  power  for 
compressing  gas  will  be  the  least  when  the  initial  tem- 
perature is  the  lowest,  and  it  will  be  shown  later  on  that 
cooling  before  compression  will  effect  a  considerable  sav- 
ing, if  the  gas  to  be  compressed  is  drawn  from  the  holder 
exposed  to  the  sun,  provided,  of  course,  that  cooling 
water  may  be  had  at  a  small  expenditure  of  power. 

Problem — To  find  the  volume  immediately  after  com- 
pression. 

V 
Consult  Table  11,  and  under  the  headingT^  and  oppo- 

P 
site  the  pressure  ratio  --    the  proper  value  will  be  found; 

and  it  must  always  be  remembered  that  these  Values  of 
temperature  and  volumes  assume  no  radiation  of  heat 
whatever,  for  when  the  heat  generated  by  compression 
has  radiated  the  temperatures  and  volumes  are  as  cal- 
culated Isothermally, 

V 
Please  note  that  ^^  is  measured  from  the  end  of  the 


THE  COMPRESSION   AND  TRANSMISSION   OF  GAS. 


27 


Table  JJ           /ioiAo/iT/c  T/{bl£  ^o/r cjis             July  /303 

T 
To 

^-' 

Vo 

JV<//^ss/r 

D/rr 

y/ufiB£/r 

Oifr. 

/£ 

/0¥66 

■*/£ 

04^e 

3063 

300 

/•^ 

/oa7a 

36S 

037S 

7763 

733 

/6 

//^-^  7 

336 

■/£'4'7 

7030 

334 

JS 

//J^B3 

309 

/3S3 

6436 

433 

£ 

/  /S3£ 

£37 

/33£ 

334-a 

-4/3 

£2 

/  £/73 

£63 

£/73 

3333 

332 

£-t- 

/  £^^^7 

£S/ 

£^4-7 

3/36 

30/ 

£6 

/■££,3e 

£6>a 

■  £4,38 

■4336 

£64 

^S 

/£366 

y33 

£366 

462/ 

£33 

3 

/3/e/ 

£/'¥■ 

3/6/ 

4333 

£07 

3£ 

/  337S 

£0^ 

3373 

4/3/ 

736 

3¥- 

/  3S73 

73  S 

■3373 

3333 

ysa 

36, 

/377'^ 

/aa 

3  77^ 

33£7 

73£ 

JS 

/336£ 

/ao 

336Z 

■3673 

739 

^ 

/^/'¥-£ 

773 

>4/<i£ 

■3336 

7£€ 

'^£ 

/■^3/S 

760 

■'^J/S 

■34/0 

//7 

-J*  -St 

/-f-f^33 

/6Z 

■-f433 

■3£33 

70s 

-^e 

/-t^^'fS 

73-7 

-fS-f-S 

3/as 

33 

'fS 

/  -^ao^ 

/S£ 

^^ao£ 

■3036 

34 

s 

/^3S^ 

637 

43S^ 

£39Z 

333 

6 

ys6S/ 

6/S 

ses/ 

■£603 

£34 

7 

/■6266 

3S£ 

a-£6e 

£323 

£22 

a 

/6S/^ 

S03 

63/3 

■£/03 

777 

3 

/■Z3£/ 

■*6£ 

■732/ 

■73£6 

747 

/o 

/  77S3 

7IS3 

/773 

7:,    Trx. 

28  THE  COMPRESSION   AND  TRANSMISSION   OF   GAS. 

Stroke.     The  difference  given  in  Table   1 1    will   enable 

p 
greater  or  lesser  values  of  ^  to  be  conveniently  deter- 
mined by  simple  rules  of  proportion. 

From  this  Table  the  adiabatic  curve  can  be  readily 
drawn. 

Refer  to  Table  i,  Figure  2. 

lyCt  A  B  be  the  intake  line  and  CD  the  line  of  o pres- 
sure, these  lines  representing  the  piston  stroke.  Divide 
A  B  into  a  decimal  scale,  beginning  at  B  erect  F  D  2X 
the  end  of  the  stroke  and  divide  it  into  equal  values  q{  B  D 
and  B  D  may  be  the  value  at  sea  level  or  at  an  altitude 
or  It  may  be  any  intake  pressure  whatever,  these  rules 
will  always  apply.  These  values  oi  B  D  may  be  sub- 
divided into  five  parts,  where  special  accuracy  is  requied, 
and  their  values  will  also  be  found  in  Table  11. 

P 
D  //representing  a  ratio  of  _^  =  2,  the  corresponding 

V 
value  of  j^^   will  be  found  in  Table  11  to  be  .5948,  and 

laying  off  the  value  the  point  ►S"  will  be  found. 

P  V^ 

Similarly  at  G  representing  p^  =  3  we  find y^^  =  .4388, 

and  laying  this  off  we  find  that  the  point  M.     And  then 

P  V 

F  representing  ~  =  4  has  a  value  for     ^^  of  .3536,  and  we 

lay  off  this  value  and  find  point/.     Joining  the  point  J 

M  S  A  WQ^  develop  the  adiabatic  curve,   and  the  shape 

of  this  curve  will  depend  upon  the  length  of  the  card,  the 

C 
value  of  -,-    or  y.     The  equation  of  the  curve  is  P   V^ 

—  p'  yy  or  referring  to  the  diagram. 

MOxiMGY  =/Lx(/n' 

Problem — To  determine  the  power  to  compress  a  gas 
adiabatically. 

All  that  precedes  this  subject  has  been  necessary  to  its 
proper  understanding,  and  while  possible  the  various 
symbols  are  well  remembered,  it  will  probably  be  better 
to  group  them  together,  so  that  they  may  be  readily 
referred  to. 


THE  COMPRESSION   AND  TRANSMISSION   OF  GAS.  29 

P°  is  always  the  lesser  absolute  pressure,  and  conse- 
quently the  intake  pressure  in  compression.  We  shall 
take  this  as  14.7  at  sea  level,  for  the  4-inch  water  press- 
ure of  the  gas  will  not  fill  the  cylinder  at  any  greater 
than  atmospheric  pressure.  P  is  the  final  absolute 
pressure. 

T"  is  the  initial  absolute  pressure,  and  unless  other- 
wise specified  is  taken  at  60°  Fah.,  or  520°  absolute. 
That  temperature  being  the  probable  temperature  of  the 
gas  mains. 

T  is  the  final  absolute  temperature. 

V  is  the  volume  at  P° . 

V  is  the  volume  at  P. 

P'  V  T'  are  intermediate  pressures,  temperatures  and 
volumes. 

L  is  the  work  expressed  in  foot-pounds. 

H Pis  horsepower. 

M E  P  IS  mean  efi'ective  pressure,  which  is  always 
gauge  pressure. 

W  is  the  weight  of  a  unit  volume  or  one  cubic  foot  of 
our  standard  gas  at  60°  Fah.  and  at  sea  level,  with  an 
absolute  pressure  of  14.7  lbs.  per  square  inch,  or  2116.8 
pounds  per  square  foot,  and  equals  .03063  pounds  avoir- 
dupois. 
J  is  Joules'  equivalent  taken  at  772  foot-pounds. 

C"  is  the  specific  heat  at  constant  pressure  =  .6884. 

C  is  the  specific  heat  at  constant  volume  =  .5159. 

y  IS  —=1.334. 
y  y-i  I 

y-i    ^      y        ^  y     '"" 

J  C''=772  X  .6884=531.45  foot-pounds. 

These  two  values  are  Joules'  equivalent  for  i  lb.  of 
gas. 

/  W  C^  =  531-45  X  03063  =  16.28  foot-pounds  = 
Joules'  equivalent  for  i  cubic  frot  of  gas. 

/  W  O'  T"  =  16.28  X  520  =  8465  foot-pounds. 

-^  X  /»"  F  =  4  X  144  X  I  X  14.7  =  8465.  = 
y-i 

the  intrinsic  energy  of  i  cubic  foot  of  gas  at  60°  Fah.,  or 


30  THE   COMPRESSION   AND   TRANSMISSION   OF   GAS. 

to  reduce  those  values  of  foot-pounds  to  horsepower,  we 
have 

J  WC  r  =  J465  _    2564  HP 

y-\  33000 

All  of  these  foregoing  quantities  are  constants  to  be 
used  in  determining  the  power  to  compress  gas,  ard  as 
we  have  said  before,  are  all  based  on  a  quantity  of  i 
cubic  foot  of  our  standard  gas  at  sea  level  and  60°  Fah. 
We  mentioned  at  the  beginning  of  this  paper  that  the 
power  to  compress  any  gas  might  be  expressed  by  the 
general  formula 

L=JWO'{T—  T"),  or  to  put  it  in  another  form, 

L=/ivo^r<^('^^-  i) 

You  now  at  once  recognize  the  prefix/  IV  O'  T^  as 
the  one  for  which  we  have  found  a  value  of  8465  foot- 
pounds.    Therefore,  for  our  standard  gas  we  have 

L  =  8465    (     ^,  ~  I  )  which  is  a  practical  formula. 

T 
You  also  recognize  that  -^  is  all  you  need   solve,    and 

these  values  are  all  given  in  Table   1 1   for  the  various 

P 
values  of -p^.  We  can  now  understand  our  first  problem. 

How  many  foot-pounds  are  necessary  to  compress  i  cu. 
ft.  of  our  standard  gas  to  14.7  pounds  gauge  pressure  ? 

-p-  —  — ^  =  2.       Consulting    Table    1 1     we  find 
P"       14.7 

T 

^=^  —  I  =  .1892  and  8465  X  .1892  =  1601.57  foot-pounds, 

and  the  same  method  may  be  applied  for  all  pressures. 
If  we  use  the  value  of  /  IV  C"  T"  in  horsepower,  we 

we  have  //P  =  .2564  (w~i)  ^  perfectly  practical  for- 
mula for  I  cu.  ft.  of  our  standard  gas  at  60°  Fah.  and  at 
sea  level. 

Our  previous  example  would  then  be  rendered: 

L  =  .2564  X  .1892  =  .0485  hp.   for  I  cu.  ft.  com- 
pressed to  14.7  lbs  gauge.     At  80  lbs.  gauge  pressure. 


THE   COMPRESSION   AND  TRANSMISSION  OF  GAS.  31 

P  T 

-^  =  6.442  and  ^.,  -1.593 

H P  =  .2564  X  .593  =    .1520  hp,   per  cu.    ft.,  or 
15.20  HP  per  100. 

MEAN   EFFECTIVE    PRESSURES. 

It  will  be  found  that  inasmuch  as  we  learn  from  an 
indicator  what  our  gas  compressor  is  doing,  and  inas- 
much as  J/ £"/*  pressures  are  quickly  determined  by  a 
planimeter  from  an  indicator  card,  that  to  become 
familiar  with  what  the  M  E  P  should  be  and  compare  it 
with  what  the  compressor  is  doing  is  the  best  practical 
way  of  dealing  with  the  subject. 

We  found  that 


and  that 


L=JWO'T^(^^  -  i) 

J  WO-  T"  =^   -^   P" t/\  therefore 

y-i 

L  =  -^  P"  V  (~  -  I  ) 
y-i  \  /  °  / 

L  must  always  equal  M  E  P  X  F°,  we  have 
MEPX    yo  =  _JL^  po   yo  /Zl  _   I  )  or 

ME  P=  -^—  P« r^^  -  I  ")  and  since 


y 

=  4,  we  have  for  our  standard  gas 


y-i 
MEP=  ^P-{^^-  i) 

Take  80  lbs.  gauge  pressure. 

T 

-7pr^  —  I  =  .593  as  determined  in  a  former  example 

by  Table  11. 

P-  =  14.7 

ME  /•  =  4  X  ,593  X  14.7  =  34.86  lbs.  per  sq.  in. 

For  our  standard  gas  for  one  cu.  ft. 

ME  P  =  ^y.  14.7  (l^o  -  I  )  =  58.8  (f-o  -  I  ) 

..   ^  1±±X   iX  M  E  P  ,^     »^ri   r) 

H  P  =  ~^-^ =  .00436  X  ME  P,OT 

33000 

HP  =  .00436  X  58.8  (^o  -  I  )  =-2564  (y-o  -  1  ) 
the  same  result  we  obtained  in  a  former  example. 


32  THE  COMPRESSION   AND  TRANSMISSION   OF  GAS. 

INITIAL  TEMPERATURES. 

The  general  expression  for  the  work  of  compression 
being 

T 
it  is  evident  that  so  long  as  v^-    remains    constant,     the 

power  to  compress  one  cubic  foot  of  the  same  gas  is  con- 
stant, but  inasmuch  as  the  temperature  of  the  mains  is 
practically  constant  and  about  60°  Fah.,  if  our  initial 
temperature  from  the  holder  should  happen  for  any 
reason  to  be  100°  Fah.,  as  it  was  entering  the  compressor, 
it  is  evident  that  the  compressor  must  make  an  extra 
number  of  revolutions  to  deliver  a  fixed  quantity  into 
the  mains  at  60°  Fah.  than  it  would  if  the  mains  were 
the  same  temperature  as  the  gas  in  the  holder,  and  the 

ratio  would  be  as  the  absolute  temperature  or ,   or   8 

520 

per  cent,  additional.  In  a  plant  where  250  horsepower  is 
used  in  compressing  the  gas,  this  would  mean  a  saving 
of  20  horsepower.  By  passing  the  gas  through  a  cooler 
before  it  reached  the  compressor  would  correct  the  loss. 
Inasmuch  as  little  water  is  required  for  this,  and  the 
water  is  in  no  wise  impaired  for  other  purposes,  that  this 
cooling  could  always  be  done.  ]^ice  versa,  if  the  tempera- 
ture of  the  holder  was  lower  than  the  mains,  as  in  winter, 
there  would  be  a  corresponding  gain  and  some  of  the 
otherwise  lost  heat  of  compression  would  be  utilized  in 
expanding  the  gas  to  a  temperature  corresponding  to  the 
main.  In  the  long  run,  the  gain  might  balance  the  loss, 
if  no  cooling  were  done,  but  it  seems  a  business  proposi- 
tion to  save  where  possible,  especially  where  it  costs 
little  or  nothing. 

TWO    STAGE    COMPRESSION. 

If  we  consider  the  general  equation  for  the  work  per- 
formed in  compressing  any  gas, 
L^  J  W  a  {T—  r«) 
we  note  that  the  only  variable  is    T,  the  final  tempera- 
ture, if  our  initial  -temperature  remains  the  same.     In 
other  words,  the  diflfcrence  between  the  initial  and  fin 


THE  COMPRESSION   AND  TRANSMISSION   OF  GAS.  33 

temperature  determines  always  the  power  expended  in  a 
compressor,  just  as  it  does  the  power  given  out  by  any 
heat  engine.  It  is  evident,  then,  that  the  lower  we  keep 
the  final  temperature  the  less  power  it  takes.  Water 
jacketing  the  cylinders  accomplishes  but  little,  probably 
from  3  to  5  per  cent.,  for  the  reason  that  gases  being  such 
poor  heat  conductors  that,  while  they  are  rapidly  drawn 
in  and  pushed  out  of  the  compressing  cylinder,  there  is 
not  the  time  for  the  heat  to  radiate  through  the  cylinder 
walls,  and  only  the  portion  immediately  in  contact  with 
the  cool  cylinder  walls  suffers  any  reduction  of  tempera- 
ture. The  water  jacket  keeps  the  cylinder  walls  cool  so 
that  lubrication  is  effective  and  is  valuable  for  that  reason 
principally. 

Practically  speaking,  the  compression  is  adiabatic,  or 
even  greater  because  the  pressure  in  the  cylinder  is 
always  greater  than  the  receiver  on  account  of  the  work 
expended  in  forcing  the  work  through  the  valve  open- 
ings, and  this  extra  heat  generated  overruns  the  adia- 
batic temperature  corresponding  to  the  receiver  pressure. 

The  water  jacket  beirg  ineffective,  the  device  of  stage 
compression  was  inaugurated,  where,  after  the  gas  was 
compressed  to  a  portion  of  the  final  pressure  'in  a  cylin- 
der, it  was  discharged  into  an  intercooler,  its  tempera- 
ture reduced  to  the  initial  and  then  compressed  by  a 
smaller  cylinder  to  the  final  pressure.  The  work  was 
found  to  be  a  minimum  when  the  final  temperature  of 
each  stage  was  the  same. 

If  we  represent  the  initial  pressure  by  P^  and  the  final 
by  P',  and  volumes  and  temperatures  similarly,  we  shall 
have,  using  our  general  formula  for  work  expended, 

L  =  — - —  P°  V  (  y^  —    \\  for  first  stage  and 
y-i  \  j"  / 

V  /T'  \ 

L'  —  — =—  P  V  I  ,A  —  I   I  for  second  stage. 
y-i  \r  / 

We  know  that  before  compression  /*"  T'"  must  equal 

P  y,  consequently  if/,  is  desired  to  equal/.',  we  must  have 

(^„  - 1 )  =  (f  -  ■  > 

ZL  =  H 


or 


34  THE  COMPRESSION   AND  TRANSMISSION   OF   GAS. 


G^f 


Y  ^  \p~)^  '  °^  reducing 


7"        />'  

jj-„  =  -p  OT  P'  =  P"*  P' or  P  =  v//^°  7^' 

In  other  words,  to  make  the  work  in  two  stages  equal, 
and  to  have  ihe  work  a  minimum,  P,  the  intermediate 
pressure,  must  be  a  mean  proportional  between  the  ini- 
tial and  the  final  pressure,  the  volumes  and  the  piston 
areas  must  follow  the  same  law,  since  we  naturally  make 
make  the  strokes  alike. 

For  an  example,  let  us  take  80  lbs.  final  gauge  pressure: 

P  =  y/ P^  P'  or  V14.7  X  94.7   =   37.31   absolute 

22.61  gauge  pressure.     This  makes 

^"       37-31 

P    ^  iW  =  '-54'  ^"d 

P'  ^  94.7 

P        37T1  "  '-^^ 
and  inasmuch  as  these  pressure  ratios  are  the  same,  the 
work  expended  on  each  stage   will  be  the  same  and  the 
piston  ratio  will  be  2.54  also. 

We  found  for  the  standard  gas  that 

/r/>  =  .2564(|-„-  i) 

Referring  to  Table  11 ,  we  find  when 

P  .       T  . 

pZ  =  2.54,  that  —  =  1.2624 

Then  H P  ^  -2564  X  .2624  =  .06727  for  each  stage 
and  for  both  stages,  2  X  .06727  =  .13454  HP. 

It  will  be  remembered  that  we  calculated  the  single 
stage  HP  for  80  lbs.  in  a  former  example  as  .1520.  We 
have  then  13.45  H P  p&r  100  cu.  ft.  against  15.10  H  P^ 
a  saving  of  13  per  cent,  in  power. 

If  the  maintaining  of  a  low  temperature  is  any  advan- 
tage in  gas  compression,  we  have  a  temperature  of  366° 
Fah.  in  the  single  stage  compression  against  195°  Fah. 
in  the  two  stage,  a  remarkable  difference.  Suppose  now 
that  we  have  a  cylinder  having  an  area  of  100  sq.  inches, 
when  we  compress  to  80  lbs.  single  stage  the  maximum 
strain   is   8000   lbs.,   if  the   compressor   is  single   stage 


THE   COMPRESSION   AND  TRANSMISSION   OF  GAS.  35 

and  4522  lbs.  if  the  compressor  is  a  tandem  compound,  a 
remarkable  difference,  tending  to  show  that  we  can  build 
the  compressor  very  much  lighter  for  the  same  work. 

Another  point  in  favor  of  the  two-stage  compressor,  it 
has  a  greater  volumetric  efficiency.  A  piston  never 
delivers  from  a  cylinder  an  u mount  of  gas  equal  to  its 
displacement,  because  clearance  spaces  are  filled  with 
gas  at  the  discharge  pressure,  which  expands  in  the 
return  stroke  of  the  piston  and  occupies  more  or  less 
space  according  to  the  ratio  of  compression  and  the 
amount  of  clearance.  The  greater  the  temperature  of 
compression,  the  hotter  the  piston  and  heads  and  valves 
get,  and  the  less  weight  of  gas  enters  the  cylinder  on 
account  of  its  expansion.  There  are  other  losses  which 
need  not  be  mentioned  here,  but  these  two  are  sufficient 
to  make  the  volumetric  efficiency  of  single  stage  com- 
pressors at  80  lbs.  average  about  75  per  cent. 

It  will  be  readily  seen  that  the  initial  cylinder  of  a  two 
stage  machine  at  80  lbs.  will  have  its  Clearance  losses 
divided  by  2.54,' because  that  will  be  the  relative  ratio  of 
pressures  and  the    temperature  losses  in   proportion  to 

— ~  because  that  is  the  temperature  ratio. 

These  combined  will  make  the  average  two  stage 
compressor  good  for  90  per  cent,  volumetric  efficiency — 
in  other  words,  15  per  cent,  better  than  a  single  stage. 
One  can,  therefore,  affi^rd  to  pay  at  least  15  per  cent, 
more  for  a  two  stage  machine  than  for  a  single  stage 
machine,  the  intake  cylinders  being  the  same  size,  and 
this  ettra  15  per  cent,  will  nearly,  or  sometimes  quite, 
pay  for  the  difference  in  price. 

It  is  evident  from  the  calculations  we  have  made  that 
the  efficiency  of  a  two  stage  machine  over  the  single 
stage  increases  directly  as  the  pressure  ratios  increase, 
and  inasmuch  as  altitude  increases  pressure  ratios,  it  is 
evident  that  the  higher  the  altitude  the  more  urgent  be- 
comes the  necessity  lor  using  the  two  stage  machines, 
and  at  altitudes  above  3000  feet  it  is  practically  impera- 
tive. 

Theoretically,  an  infinite  number  of  stages  would  give 
isothermal   compression,    but  practically    the  losses  in- 


36 


THE   COMPRESSION    AND   TRANSMISSION   OF   GAS. 


volved  in  driving  the  gas  through  too  many  cylinders 
and  valves  would  offset  this  gain,  and  we  can  consider 
that  two  stages  will  probably  be  the  limit  for  all  ordinary 
purposes. 

ALTITUDE   COMPRESSION. 

We  found  that  it  took  the  same  power  to  compress  one 
cubic  foot  of  gas  at  any  temperature  to  the  same  final 
pressure,  provided  the  initial  pressures  were  the  same, 
and  it  naturally  followed  that  it  took  more  power  to  com- 
press the  same  weight  at  higher  temperatures,  because 
there  would  be  a  larger  volume  and  the  piston  would 
have  to  make  more  strokes. 

Altitude  acts  like  an  increase  of  temperature  in  lessen- 
ing the  density  of  a  gas,  but  it  introduces  another  ele- 
ment, viz.,  change  of  initial  pressure,  so  that  as  we 
reach  higher  altitudes  the  pressure  ratio  is  constantly 
increasing,  which  means,  of  course,  that  the  tempera- 
tures of  compression  is  increasing  and  more  work  per 
unit  of  gas  weight  is  being  done,  but  the  weight  is  con- 
stantly decreasing  as  we  ascend,  and  the  combination  of 
these  results  is  that  while  it  takes  less  work  to  discharge 
any  given  cylinder  full  of  gas  at  an  altitude,  the  in- 
creased number  of  strokes  necessary  to  compress  a  weight 
equivalent  to  a  given  sea  level  volume  is  considerably 
greater. 

Table  17.  July,  1905. 


Altitude. 

P 

r 

P" 

yo 

T 

0  <u    ■ 

0  "£  0 
o-ao 

Kquivalent     to 
produce  same 
compressed 
gas  at  altitude 

V. 

3 
to 

p< 

.2 

c 

to 

a; 
u 

0. 

r— I 

3 

a 
0 

Sea  level 
io,cxX)  ft. 

6.44 

9-47 

4 

4 

14-7 
10. 

1X144 
IX 144 

■593 
•753 

5020 
4337 

5020 
6375 

147 
10. 

80 
80 

T=  390°  Fah.  at  sea  level. 

T  =  450°  Fah.  at  10,000  feet  altitude. 

Table  17,  shows  a  comparison  between  compres- 
sing gas  at  sea  level  and  at  10,000  feet  altitude.  The 
columns  3  to  6,  inclusive,  comprise  the  components  of 
the  general  formula  for  compressing  gas,  and  it  is  inter- 
esting to  note  the  variable  quantities.     It  will  be  «?een 


TTIK  ei)\IP  IKSSIOX    AXD   TRANSMISSION   OF   GAS. 


37 


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38 


THE   COMPRESSION    AND   TRANSMISSION    OF   GAS. 


THE   COMPRESSION    AND  TRANSMISSION  OF  GAS.  39 

that  while  one  cubic  foot  of  the  altitude  gas  requires  less 
power,  the  increased  volume  necessary  to  produce  a  com- 
mon result  makes  it  require  25  per  cent,  more  power. 

It  will  also  be  noted  that  the  final  temperature  is  quite 
high  in  comparison  to  sea  level  compression,  which 
speaks  loudly  for  two  stage  compression. 

FLOW   OF    GAS    IN     PIPES. 

After  reading  the  report  of  the  committee  on  "The 
Flow  of  Gas  in  Pipes,"  for  the  Ohio  Gas  Light  Associa- 
tion, as  published  in  the  American  Gas  Light  Journal^ 
April  24,  1905,  the  general  impression  would  be  that  the 
formulae  were  not  sufficiently  reliable  to  be  of  great 
service,  because  there  was  a  variation  in  the  results  of  a 
given  problem  of  from  i  to  200  per  cent.  It  would  seem, 
however,  that  six  formulae  out  of  the  nine  do  not  vary 
15  per  cent.,  and  the  three  most  frequently  used  do  not 
vary  2^  per  cent. 

If  we  should  accept  the  largest  of  these  three,  called 
the  Pittsburg  formula,  we  would  probably  not  be  far 
wrong,  and  particularly  as  the  results  do  not  differ 
greatly  from  those  obtained  by  using  Cox'  computer,  and 
I  am  informed  by  those  who  have  used  the  computer  that 
it  is  perfectly  safe. 

Again,  the  variation  in  the  areas  of  those  pipe  sizes 
most  likely  to  be  used  are  much  more  considerable  than 
the  variations  of  any  of  the  six  formulae  above  referred 
to.  Thus,  taking  the  commercial  sizes  of  pipe  from  i" 
to  6",  the  average  variation  between  the  areas  of  each 
size  is  35  per  cent. 

If  we  therefore  make  a  practic*'  of  using  the  pipe  that 
is  the  nearest  size  larger  than  our  calculations,  we  shall 
have  an  ample  safety  factor. 

For  air  we  have  been  using  a  formula  developed  by 
Mr.  J.  E.  Johnson,  Jr.,  and  published  in  the  American 
Machinist  ]n\y  27,  1899.     (Table  17,  No.   2) 

P'  =  absolute  initial  pressure 
P"  =  absolute  final  pressure. 
Q  =  free  air  equivalent  in  cubic  feet  per  minute. 


40  THE   COMPRESSION    AND   TRANSMISSION   OF   GAS. 

L  =  length  of  pipe  in  feet. 
d  =  diameter  on  pipe  in  inches. 
Practical  results  from  this  formula  show  that   it  is   a 

little   too   liberal,    and   that  P"  —  P'"   =   i^°-^^|,^^ 

d^ 

would  be  nearer  the  results. 

The  Pittsburg  gas  formula  reduces  to  the  same  value 
when  the  proper  substitutes  are  made  for  the  relative 
specific  gravities  of  gas  and  air. 

Inasmnch  as  the  specific  gravity  of  gas  is  always  re- 
ferred to  air  as  i,  it  seems  right  that  our  gas  formula 
should  refer  to  air  and  a  coefficiency  used  for  each  gas. 

The  velocity  of  different  gases  through  a  pipe  vary  in- 
versely as  the  square  root  of  their  densities,  or  what 
amounts  to  the  same  thing,  their  specific  gravities  or 
weights  compared  to  air,  then  the  velocities  will  vary  as 


j: 


or  y/G 


Where  G  is  the  specific  gravity  of  the  gas. 
Prefixing  this  to  our  original  equation,   we  have  in 
general,  for  any  gas, 

p.,  _  p,H  ^     0005  y/G   X  %^ 

d" 
Or  -^v      

Q  =s/—cr  s z 

Inasmuch  as  certainly  for  some  considerable  time 
crude  oil  gas  will  be  most  extensively  used  by  members 
of  this  Association,  let  us  substitute  in  the  above  formula 
the  value  of  the  largest  probable  specific  gravity,  viz., 
.49,  and  we  have\/  .^^  -  .7  and 


/"^-/^"%  .00035-^  {A 


Q  =  6^^^-  ^"  X  ^' 


L 

Q  is  in  cul)ic  feet  per  minute  rather  than  per  hour,  be- 
cause all  compressors  are  so  rated. 

Table  12  give's  values  of/"^  —  P"'- iox  100  feet  for 
various  sized  pipes  and  quantities  will  be  found  conveni- 
ent for  figuring  gas  flows  in  pipes.  The  values  are  calcu- 
lated from  equation   (/?). 


THE  COMPRESSION   AND  TRANSMISSION   OF  GAS.  41 

Example  I — 

1000  cubic  feet  per  minute  of  gas  at  90  pounds  gauge 
pressure  is  discharging  into  a  4"  pipe  26; 000   feet  long. 
Required  the  terminal  pressure. 
P''  ^  10962.   (Table  13) 
P'l  _  p'n  ^  2^  o^  for  100  feet.     (Table  12) 

Multiplying  by  260  for  26,000  feet 

P'-  —  P"'^()\\0. 

P"i  ^  p'i—{^p'i  _  p"i-)  ^  10962  —  9110  =  1852. 
P'"  =  1852.     P"  =  28  pounds. 
Example  II — 

A  pipe  line  3"  diameter  and  1 1 ,000  feet  long.  Required 
to  find  the  quantity  of  gas  that  will  be  delivered  at  a 
terminal  pressure  of  ^  pound,  the  initial  pressure  being 
40  pounds. 

P"  =  2992  (Table  13) 
P'"  =  279  (Table  13) 
P'^  —  P"'  =  2713  for  11,000  feet  of  pipe  or  24.6  for 
100  feet. 

Referring  to  Table  12,  we  find  value  of  23.70  for  420 
cubic  feet  per  minute 

Example  III — 

A  pipe  line  is  11,000  feet  long  and  4"  diameter.  The 
equivalent  of  1000  cubic  feet  is  wanted  at  the  end  of  the 
line  at  10  pounds  pressure.  What  must  be  the  initial 
pressure? 

pr,_  p„,  ^  ^^  Q^  foj.  jQQ  fggj.  (Table  12).  Multiply- 
ing by  no  we  have 

P''  —P"'  =  3854  for  11,000  feet. 

P"'  =    610     (Table  13) 

pr,  ^  p".  ^  (/>'._/.".)  ^  3854  X  610=    4464. 

P'  =  v/4464- 
Referring  to  Table  13  we  find  52    pounds  gauge  pres- 
sure to  be  the  initial  pressure. 

Example  IV — 

The  equivalent  of  200  cubic  feet  per  minute  is  to  be 
put  through  a  pipe  53,000  feet  long.     The  initial  pres- 


42  THE  COMPRESSION   AND   TRANSMISSION   OF  GAS. 

sure  is  20  pounds.  The  final  pressure  must  be  6  pounds. 
What  will  be  the  size  ol  the  pipe? 

P"  =  1204  (Table  13) 

P'"  =  428  (Table  13) 

P'^  —  P"'^  =  776  for  53,000  feet  ot  pipe  or  1.464 
per  100  feet.  Referring  to  Table  12,  we  find  4"  to  be 
the  proper  size. 

SOME  CORROBORATIONS. 

Table  15  gives  at  Figure  i  a  card  from  the  gas  cylin- 
der of  a  compressor  at  Fresno,  compressing  crude  oil  gas 
at  a  pressure  of  27  pounds  gauge. 

If  we  draw  the  line  of  27  pounds  pressure  and  take  the 
M  E  P  with  a  planimeter,  following  the  curve  A  B  and 
the  straight  lines  B  C — C  D  qnd  A  B,  we  shall  have  the 
J/ £■ /*  of  a  perfect  card  following  the  actual  compres- 
sion line.  This  M  E  P  vft^  find  to  be  17.4  pounds,  using 
the  J  which  we  found  for  Fresno  gas,  the  adiabatic  M  E  P 
for  27  pounds  =  17.58,  making  a  good  check  on  our 
values. 

Figures  2  and  3  are  from  a  compressor  pumping 
natural  gas  at  Anderson,  Indiana,  each  having  an  intake 
pressure  of  1 1  pounds — drawing  lines  of  50  pounds  pres- 
sure at  Figure  2  and  60  pounds  at  Figure  3,  and  taking 
the  M  E  P  \\\  the  same  way  that  we  did  in  Figure  i,  we 
find  that  the  M E  P  iox  Figure  2  is  26  pounds,  and  for 
Figure  3,  30  pounds. 

Using  the  value  oiy  which  we  developed  for  natural 
gas  and  calculating  the  adiabatic  M  E  P,  we  find  they 
are  26.30  and  30.85  pounds,  respectively,  a  very  satis- 
factory check,  and  from  these  we  may  fairly  conclude 
that  our  theories  and  formulae  are  reasonable. 

It  will  be  noted  that  the  line  of  the  compressor  curve 
is  very  near  the  adiabatic.  even  though  the  compressors 
were  making  but  60  to  70  revolutions  per  minute.  An 
air  card  would  show  at  least  double  the  separating  space. 

This  would  appear  to  show  thatthejackets  were  doing 
but  very  little  good,  and  possibly  because  illuminating 
gas  may  be  a  much  poorer  conductor  of  heat  than  air. 

The  line  of  compression  comes  so  near  the  adiabatic 
that  we   may    well    call    the   compression  adiabatic  for 


THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 


43 


Table /4     /a/djcatoj?  CAfiDS  r/fOM  CAs  COMF/fEssoffS.  July /SOS 


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44  THE   COMPRESSION   AND   TRANSMISSION   OK  GAS. 

safety  in  our  calculations — but  while  the  M  E  P  adiaba- 
tic  for  any  pressure  represents  the  greatest  possible 
power  required  to  compress  a  gas,  a  still  greater  power 
must  be  applied — for  example  look  at  the  Fresno  card, 
Figure  i,  Table  14 — the  area  above  the  27  pounds  line 
represents  work  done  in  overcoming  the  inertia  of  the 
outlet  valves  in  pushing  the  gas  into  the  main,  and  this 
area  will  be  greater  or  less  depending  upon  the  valve 
area  and  the  size  of  the  discharge  openings  and  the  pis- 
ton speed.  It  will  also  be  noted  that  there  is  an  area 
representing  suction  work  below  the  line  A  D,  notwith- 
standing that  the  gas  has  a  4"  water  pressure  at  holder. 
This  probably  indicates  that  the  pipes  from  the  holder  to 
the  compressor  are  too  small. 

Now,  if  we  run  a  planimeter  over  the  actual  area  of 
the  card,  we  find  that  the  real  ME  P  is  19.4,  or  about 
10  per  cent,  greater  than  the  adiabatic,  and  this  agrees 
quite  well  with  ordinary  air  practice,  where  a  safe  rule 
for  single  stage  work  is  to  take  the  M  E  P  at  10  per 
cent,  above  the  adiabatic  and  the  two  stage  M  E  P  the 
same  as  the  adiabatic.  Slow  speed,  well  constructed 
compressors  will  do  somewhat  better,  but  it  is  well  to 
calculate  on  the  average  type. 

Now,  for  brake  power  to  be  delivered  to  a  gas  com- 
pressor, we  have  to  allow  a  mechanical  efficiency  of  the 
compressor  at  not  to  exceed  85  per  cent.,  so  that  this  15 
per  cent,  loss  combined  with  the  10  per  cent,  loss  in  the 
cylinder  points  to  the  fact  that  we  should  add  26^  per 
cent,  to  the  adiabatic  H  P  for  the  brake  power  required. 

The  steam  engine  cards  on  the  Fresno  compressor 
show  an  M  E  P  reduced  to  the  size  of  the  air  cylinder  of 
20.75  pounds,  or  20  per  cent,  higher  than  the  adiabatic 
air  M  E  P,  but  this  compressor  had  a  Meyer  cut-ofF, 
which  helped  its  economy  considerable. 

Referring  to  Table  9,  column  17,  gives  the  formula  for 
computing  the  power  to  compress  one  cubic  foot  of  the 
gas  at  sea  level  and  60°  Fah.  If  the  calculation  be  made 
it  will  be  noted  that  it  takes  practically  the  same  power 
to  compress  one  cubic  foot  of  any  of  these  gases,  conse- 
quently Table  19  may  be  used  generally. 


THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 


45 


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46  THE  COMPRESSION   AND  TRANSMISSION  OF  GAS. 

In  conclusion  your  attention  is  called  to  Table  19, 
which  contains  in  convenient  form  the  results  which  we 
have  obtained,  and  which  it  is  hoped  you  will  find  very 
helpful  in  considering  thermodynamic  questions  regard- 
ing the  standard  illuminating  gas  made  from  crude  oil. 


517345 


UNIVERSITY  OF  CALIFORNIA  UBRARY 


K'., 


JG 


tJt'V-' 


